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Scaling Laws for Polar Integrals

Derive the polar integral scaling law and use homogeneity to predict how an integral changes when its region is dilated.

Intermediate 10 min read

What You'll Learn

  • Describe a dilated planar region using scaled polar radial bounds
  • Derive the scaling factor contributed by the polar area element
  • Apply the scaling law to homogeneous integrands
  • Separate the scaling of a sum of integrands with different degrees
  • Verify scaled integrals by evaluating their polar bounds directly

Why Scaling Changes an Integral

A dilation enlarges or shrinks every length in a region by the same positive factor. The angles stay the same, but the area changes: doubling every length multiplies area by four. A double integral can change by more than the area factor, because the integrand may also change when its input points move farther from the origin.

Polar coordinates make both effects visible. The area element is \(r\,dr\,d\theta\), so scaling the radial coordinate scales the area element by the square of the dilation factor. If the integrand is homogeneous, its own scaling is another predictable power. Combining these two contributions gives a useful rule for evaluating integrals on enlarged or reduced regions.

Dilating a Region in Polar Coordinates

For a positive number \(\lambda\), the dilation of a planar region \(D\) by factor \(\lambda\) is the region \(\lambda D\) consisting of all points \((\lambda x,\lambda y)\) with \((x,y)\in D\). The origin stays fixed, all distances from the origin are multiplied by \(\lambda\), and angles are unchanged.

Suppose, for example, that a region has a polar description with angular bounds \(\alpha\leq\theta\leq\beta\) and radial bounds \(a(\theta)\leq r\leq b(\theta)\). Its dilation has the same angular bounds. At each angle, its radial bounds become \(\lambda a(\theta)\leq r\leq\lambda b(\theta)\).

Definition: A function \(f:\mathbb{R}^2\to\mathbb{R}\) is homogeneous of degree \(p\) if \(f(\lambda x,\lambda y)=\lambda^p f(x,y)\) for every \(\lambda>0\) and every \((x,y)\) in its domain. The value \(p\) is the degree of homogeneity.

For instance, \(f(x,y)=x^2+y^2\) is homogeneous of degree \(2\), since \(f(\lambda x,\lambda y)=\lambda^2(x^2+y^2)\). The constant function \(f(x,y)=1\) is homogeneous of degree \(0\). By contrast, \(1+x^2+y^2\) is not homogeneous of a single degree: its constant term and quadratic term scale differently.

Deriving the Scaling Law

Write the polar form of the integrand as \(f(r\cos\theta,r\sin\theta)\). If \(D\) has the radial description above, its integral is expressed using the polar area element \(r\,dr\,d\theta\). For the dilated region, substitute \(r=\lambda s\), so \(dr=\lambda\,ds\). The area element then becomes \(r\,dr=(\lambda s)(\lambda\,ds)=\lambda^2s\,ds\).

$$ \begin{aligned} \iint_{\lambda D} f(x,y)\,dA &=\int_{\alpha}^{\beta} \int_{\lambda a(\theta)}^{\lambda b(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\\ &=\lambda^2\int_{\alpha}^{\beta} \int_{a(\theta)}^{b(\theta)} f(\lambda s\cos\theta,\lambda s\sin\theta)\,s\,ds\,d\theta. \end{aligned} $$

The factor \(\lambda^2\) comes from area alone. If \(f\) is homogeneous of degree \(p\), then its value at the dilated point is \(\lambda^p\) times its value at the original point. This gives the central rule.

Theorem: Suppose \(D\) is a planar region, \(\lambda>0\), and \(f\) is homogeneous of degree \(p\). If the integrals over \(D\) and \(\lambda D\) exist, then $$ \iint_{\lambda D} f(x,y)\,dA = \lambda^{p+2}\iint_D f(x,y)\,dA. $$

The same argument applies to regions described by several radial intervals, provided the integral is split over those intervals. More generally, the polar radial bounds need not be constant: multiplying each radial bound by \(\lambda\) still gives the dilation, and the substitution above applies angle by angle.

Why the Exponent Is \(p+2\)

There are two independent changes. First, the area of each small polar piece is scaled by \(\lambda^2\). Second, a degree-\(p\) integrand has its value multiplied by \(\lambda^p\) at the corresponding dilated point. The total factor is the product \(\lambda^2\lambda^p=\lambda^{p+2}\). The derivation also explains why this is not just a rule about area: the integrand’s degree matters.

For the constant integrand \(f=1\), the degree is \(0\), so the integral scales by \(\lambda^2\), as area should. For an integrand of degree \(1\), the integral scales by \(\lambda^3\). For an integrand of degree \(2\), it scales by \(\lambda^4\).

Worked Example: Scaling the Area of a Sector

Let \(D\) be the sector \(0\leq r\leq2\), \(0\leq\theta\leq\pi/3\). Find its area, then find the area of its dilation by \(\lambda=3\).

Area is the integral of \(f=1\), which has degree \(p=0\). First evaluate the original area:

$$ \begin{aligned} \iint_D 1\,dA &=\int_0^{\pi/3}\int_0^2 r\,dr\,d\theta\\ &=\int_0^{\pi/3}\left[\frac{r^2}{2}\right]_0^2d\theta\\ &=\int_0^{\pi/3}2\,d\theta =2\left(\frac{\pi}{3}\right) =\frac{2\pi}{3}. \end{aligned} $$

The scaling factor for a degree-zero integrand is \(3^{0+2}=9\). Thus the dilated sector has area \(9(2\pi/3)=6\pi\). Check directly: its radial bound is \(0\leq r\leq6\), and

$$ \int_0^{\pi/3}\int_0^6 r\,dr\,d\theta = \int_0^{\pi/3}18\,d\theta = 18\left(\frac{\pi}{3}\right) = 6\pi. $$

The direct calculation agrees with the scaling rule.

Worked Example: A Quadratic Integrand on a Disk

Let \(D\) be the unit disk and \(f(x,y)=x^2+y^2\). Find the integral over the disk of radius \(2\), using the integral over \(D\) and the scaling law.

Since \(f(\lambda x,\lambda y)=\lambda^2f(x,y)\), its degree is \(p=2\). In polar coordinates, \(f=r^2\), so the original integral is

$$ \begin{aligned} \iint_D (x^2+y^2)\,dA &=\int_0^{2\pi}\int_0^1 r^2r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^1d\theta\\ &=\int_0^{2\pi}\frac14\,d\theta =\frac14(2\pi) =\frac{\pi}{2}. \end{aligned} $$

The disk of radius \(2\) is the dilation of \(D\) by \(\lambda=2\). The integral therefore scales by \(2^{2+2}=16\), giving \(16(\pi/2)=8\pi\).

To check, integrate directly over the larger disk:

$$ \begin{aligned} \int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta &=\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^2d\theta\\ &=\int_0^{2\pi}4\,d\theta =4(2\pi) =8\pi. \end{aligned} $$

Both calculations give \(8\pi\).

Worked Example: A Degree-Two Integrand on a Quarter-Disk

Let \(D\) be the unit quarter-disk \(0\leq r\leq1\), \(0\leq\theta\leq\pi/2\), and let \(f(x,y)=xy\). Find the integral over the dilation by factor \(3\).

The integrand has degree \(2\), because \(f(\lambda x,\lambda y)=(\lambda x)(\lambda y)=\lambda^2xy\). In polar coordinates, \(xy=r^2\cos\theta\sin\theta\), so the original integral is

$$ \begin{aligned} \iint_D xy\,dA &=\int_0^{\pi/2}\int_0^1 r^3\cos\theta\sin\theta\,dr\,d\theta\\ &=\left[\frac{r^4}{4}\right]_0^1 \int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\\ &=\frac14\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2} =\frac14\left(\frac12-0\right) =\frac18. \end{aligned} $$

The scaling factor is \(3^{2+2}=81\), so the integral on the dilated quarter-disk is \(81(1/8)=81/8\). Directly, the dilated region has \(0\leq r\leq3\), and

$$ \begin{aligned} \int_0^{\pi/2}\int_0^3 r^3\cos\theta\sin\theta\,dr\,d\theta &=\left[\frac{r^4}{4}\right]_0^3 \left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2}\\ &=\frac{81}{4}\cdot\frac12 =\frac{81}{8}. \end{aligned} $$

This verifies the predicted value.

When the Integrand Is Not Homogeneous

A sum of terms with different degrees does not generally have one scaling factor. Instead, apply the change of variables to each term, or split the integrand into homogeneous parts and scale each part with its own exponent. This works because integration is linear: the integral of a sum is the sum of the integrals, when the integrals exist.

Worked Example: Scaling a Sum of Different Degrees

Let \(D\) be the annular sector \(1\leq r\leq2\), \(0\leq\theta\leq\pi/2\), and let \(f(x,y)=1+x^2+y^2\). Find the integral over \(2D\).

Split the integrand into a degree-zero term and a degree-two term. First compute each integral over \(D\):

$$ \begin{aligned} \iint_D 1\,dA &=\frac{\pi}{2}\int_1^2 r\,dr =\frac{\pi}{2}\left[\frac{r^2}{2}\right]_1^2 =\frac{\pi}{2}\cdot\frac32 =\frac{3\pi}{4},\\ \iint_D (x^2+y^2)\,dA &=\frac{\pi}{2}\int_1^2 r^3\,dr =\frac{\pi}{2}\left[\frac{r^4}{4}\right]_1^2 =\frac{\pi}{2}\cdot\frac{15}{4} =\frac{15\pi}{8}. \end{aligned} $$

Under dilation by \(2\), the constant term scales by \(2^2=4\), and the quadratic term scales by \(2^4=16\). Therefore the desired integral is

$$ 4\left(\frac{3\pi}{4}\right)+16\left(\frac{15\pi}{8}\right) =3\pi+30\pi =33\pi. $$

Check directly: the dilated sector has \(2\leq r\leq4\). Thus

$$ \begin{aligned} \iint_{2D}(1+x^2+y^2)\,dA &=\frac{\pi}{2}\int_2^4(1+r^2)r\,dr\\ &=\frac{\pi}{2}\left(\int_2^4r\,dr+\int_2^4r^3\,dr\right)\\ &=\frac{\pi}{2}\left(6+60\right) =33\pi. \end{aligned} $$

The direct evaluation agrees. Applying a single power of \(2\) to the whole integrand would not be valid, since its two terms have different degrees.

A Common Scaling Mistake

A frequent error is to include the change in the integrand but forget the change in area. For a degree-two integrand, this leads to using \(\lambda^2\) instead of \(\lambda^4\). For example, if a degree-two integral over \(D\) is \(5\), then its integral over \(3D\) is not \(3^2(5)=45\). The integrand contributes \(3^2\), and the area element contributes a further \(3^2\), so the correct calculation is \(3^{2+2}(5)=81(5)=405\).

Another useful check is the constant function. If the rule predicts that the area of a region scales by \(\lambda^4\), something has gone wrong: area corresponds to degree \(0\), so it must scale by \(\lambda^2\).

Using the Rule Reliably

Before applying a scaling law, check that the new region really is a dilation about the origin: its angles must be unchanged and all its radial distances multiplied by the same positive \(\lambda\). Then identify the degree of the integrand. If the integrand is a sum of different degrees, split it into parts instead of assigning the entire sum a single degree.

1
Identify the dilation.
Confirm that the region changes from \(D\) to \(\lambda D\) with one factor \(\lambda>0\).
2
Determine the degree.
Check how the integrand changes when both coordinates are multiplied by \(\lambda\).
3
Combine the factors.
For degree \(p\), multiply the original integral by \(\lambda^{p+2}\); for a sum of degrees, handle each part separately.
4
Check the result.
When practical, scale the radial bounds and evaluate the polar integral directly.

The scaling rule is useful whenever the same shape is considered at different sizes. It can replace a second integration with a power calculation, while also providing a check on direct computations. The exponent records two geometric facts at once: how the integrand changes and how the area element changes.

Check Your Understanding

Use the scaling rule in each question. State the degree of the integrand where needed.

  1. If \(f(x,y)=x^3\) and the integral over \(D\) is \(4\), what is the integral over \(2D\)?
  2. By what factor does the area of a region change when it is dilated by \(5\)?
  3. Is \(x^2+y^2+1\) homogeneous? If not, what degrees occur in its terms?
  4. A degree-one integrand has integral \(-3\) over \(D\). What is its integral over \(4D\)?
  5. Explain why the polar area element contributes a factor of \(\lambda^2\) under dilation.

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